# Example for Busemann function is not an exhaustion when Ricci $\ge 0$

For an open complete Riemannian manifold $M$ with non-negative sectional curvature, the Busemann function defined below is a convex exhaustion function (by Cheeger-Gromoll's proof of soul theorem)

The Buesemann function: $$b(x)=\sup_{\gamma} b_{\gamma}(x)$$ where the $sup$ is taken over all rays from a given point and $b_{\gamma}$ is the Busemann function associated with ray $\gamma$:

$$b_{\gamma}(x)=\lim_{t\to \infty}(t-d(x, \gamma(t))$$

A function $f:M\to \mathbb [a,\infty)$ is called an exhaustion function on $M$ if its sublevel set $\Omega_c:=f^{-1}((-\infty, c])$ is compact for all $c$ and $M=\cup_c \Omega_c$.

I am wondering that when we assume Ricci curvature is non-negative, is there any example where the Busemann function $b$ is not an exhaustion? (In this case the Busemann function is subharmonic.)

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What is "exaustion"? – Sergei Ivanov Oct 5 '11 at 18:56
sorry, should be exhaustion. – user16750 Oct 5 '11 at 19:01
But could you define what an "exhaustion function" is? – Deane Yang Oct 5 '11 at 19:19
Ok, what is "exhaustion"? If this means a function whose sublevels are compact (the only definition I could find), then your first claim is false. For example, consider a Euclidean space. – Sergei Ivanov Oct 5 '11 at 19:19
Sorry for any misleading due to my typo and not describe the question clearly. The Buesemann function is $b(x)=\sup b_{\gamma}(x)$ where the sup is taken over all rays and $b_{\gamma}$ is the usual busemann function. – user16750 Oct 5 '11 at 19:22

This is an interesting question. I don't have an answer but I want to clarify what is being asked as there seems to be some confusion on the issue.

Given a point $p$ and a ray $\gamma$ starting at $p$ define its Busemann function $b_\gamma(x)$ by the formula $b_\gamma(x)=\lim_{t\to\infty}(t-d(x,\gamma(t)))$. This function is known to be convex if sectional curvature is nonnegative (it's an easy consequence of Toponogov comparison). Then one can define $b=\sup_\gamma b_\gamma$ where the supremum is taken over all rays starting at $p$. If $sec\ge 0$ then the sublevel sets of $b$ are convex. If a sublevel set of $b$ were noncompact it would contain a ray starting at $p$ which contradicts the definition of $b$.

This argument doesn't work for $Ric\ge 0$ because in this case the Busemann function $b$ is only known to be subharmonic rather than convex.

So the question is whether it's still true that sublevel sets of $b$ are compact for manifolds of nonnegative Ricci curvature.

Sorry for posting this as an answer but it was too long for the comment field.

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Thanks. Do you have any idea what will the minimum set of $b$ look like? – user16750 Oct 5 '11 at 22:26
Thanks! Isn't the minimum set of $b$ always just $p$ itself? – Deane Yang Oct 5 '11 at 22:47
@Deane, no, for example, $M=S^1\times \mathbb R$, then the minimum set is $S^1$. – user16750 Oct 5 '11 at 23:17
Thanks for the counterexample. – Deane Yang Oct 6 '11 at 0:25
So the first thing I would try is to see if the Omori-Yau maximum principle could be used here. Have you (unknown, not Vitali) tried this? – Deane Yang Oct 6 '11 at 1:45

Sorry it's not an answer. I just came across a similar question recently and found a reference.

If the manifold has Ricci nonnegative and Euclidean volume growth, then the Busemann function is an exhaustion function. This is a result due to Zhongmin Shen.

See P400 Lemma 3.4 in Shen, Zhongmin, Complete manifolds with nonnegative Ricci curvature and large volume growth. Invent. Math. 125 (1996), no. 3, 393–404.

But I did not check the details of this result.

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